This set of Chemical Reaction Engineering Questions and Answers for Aptitude test focuses on “Rate Data Collection & Analysis – Method of Half Lives”.

1. State true or false.

For a second order reaction, t_{.0.5} is directly proportional to the initial concentration.

a) True

b) False

View Answer

Explanation: The general expression relating t

_{.0.5}, rate constant and initial concentration is t

_{.0.5}= \(\frac{C_{A0}^{1-n}(2^{n-1}-1)}{(n-1)k}.\) For a second order reaction, t

_{.0.5}= \(\frac{1}{kC_{A0}}.\) Hence, t

_{.0.5}is inversely proportional to the initial concentration.

2. In terms of conversion, half life is the time taken for ____

a) 50% conversion

b) 75% conversion

c) 20% conversion

d) 30% conversion

View Answer

Explanation: Half Life is the time to reduce the reactant concentration to half of initial value. Hence, 50% of the reactant is converted to product.

3. For all orders of reaction except the first order, half life is related to initial concentration as ____

a) t_{0.5} α C_{Ao}^{n}

b) t_{0.5} α C_{Ao}^{1-n}

c) t_{0.5} α C_{Ao}^{2n}

d) t_{0.5} α C_{Ao}

View Answer

Explanation: The generalised equation is t

_{0.5}= \(\frac{C_{A0}^{1-n}(2^{n-1}- 1)}{(n-1)k}.\) At half life time, for a reaction of order n, C

_{A}= 0.5C

_{Ao}.

4. The plot representing the relationship between ln(t_{0.5}) and ln(C_{Ao}) for second order reaction is ___

a)

b)

c)

d)

View Answer

Explanation: The slope of the line is (1 – 2) = -1. Hence, the straight line is linearly decrementing.

5. The relationship between t_{.0.5} and C_{Ao} for a zero order reaction is ____

a) t_{0.5} = \(\frac{1}{kC_{A0}} \)

b) t_{0.5} = \(\frac{1}{C_{A0}} \)

c) t_{0.5} = \(\frac{C_{A0}}{k}\)

d) t_{0.5} = \(\frac{C_{A0}}{2k}\)

View Answer

Explanation: For zero order reaction, \(\frac{-dC_A}{dt}\) = k. t

_{0.5}= \(\frac{C_{A0}- \frac{C_{A0}}{2}}{k}.\)

Hence, t

_{0.5}= \(\frac{C_{A0}}{2k}.\)

6. For a zero order reaction, the slope of t_{0.5} and C_{Ao} is ____

a) 2

b) 3

c) 1

d) 0

View Answer

Explanation: Slope is (1-n). At n=0, slope = 1.

7. What is the slope of the plot of ln (t_{0.5}) and ln (C_{Ao})? (Where n is the reaction order)

a) 1-n

b) 2-n

c) n

d) \(\frac{n}{2} \)

View Answer

Explanation: The plot is a straight line of slope (1-n). The equation representing the relationship is ln (t

_{0.5}) = ln\((\frac{(2^{n-1}-1)}{(n-1)k})\) + (1 – n) ln (C

_{Ao}).

8. If t_{0.5} = 100 minutes and initial concentration is 10, then the order of the reaction is ____

a) 4

b) -1

c) -2

d) 1

View Answer

Explanation: t

_{0.5}= \(C_{A0}^{1-n}.\)100 = 10

^{(1 – n)}. Hence, n = -1.

9. The fractional conversion is expressed as____

a) \(\frac{C_A}{C_{A0}} \)

b) \(\frac{C_{A0}}{C_A} \)

c) \(\frac{1}{C_{A0}} \)

d) \(\frac{C_A}{kC_{A0}} \)

View Answer

Explanation: Fractional conversion is the time required to reduce the concentration to any fraction. It is the ratio of final to initial concentration.

10. The time required to convert to a fraction of 0.7 for a first order reaction is ____

a) t_{0.7} = \(\frac{(0.7)}{C_{A0} k}\)

b) t_{0.7} = \(\frac{(-0.7)}{C_{A0} k}\)

c) t_{0.7} = \(\frac{(0.3)}{C_{A0} k}\)

d) t_{0.7} = \(\frac{(-0.3)}{C_{A0} k}\)

View Answer

Explanation: The time for converting to any fraction, F is, t

_{F}= \(\frac{C_{A0}^{1-n}(F^{n-1}-1)}{(n-1)k}.\) At F = 0.7, t

_{0.7}= \(\frac{C_{A0}^{1-2}(0.7^1- 1)}{k}.\)

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